Answer:
8/ln(9)
Step-by-step explanation:
Let u=cos(x). Then du=-sin(x) dx.
If x=0, then u=cos(0)=1.
If x=pi/2, then ucos(pi/2)=0.
The intgral is rewriten as
Integrate((9^u (-du) , from u=1 to u=0)
-9^u/ln(9) |u=1..0)
Upperlimit goes in first no matter if it is tinier in value:
(-9^0/ln(9)--9^1/ln(9))
-1/ln(9)+9/ln(9)
8/ln(9)
